STRUCTURE OF La-139
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of electromagnetic laws in favor of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for revealing the nuclear binding and the correct nuclear structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Nuclear structure of Lanthanum with 30 blank positions Naturally occurring lanthanum (La) is composed of one stable (La-139 ) and one radioactive (La-138) isotope, with the stable isotope, La-139, being the most abundant (99.91% natural abundance). Comparing the Lanthanum of 57 protons (odd number) with Barium of 56 protons (even number ) we conclude that the structure of Lanthanum brakes the high symmetry of the structure of Barium, because here there is not any additional proton (p58) to fill the 1(p) existing between n43 and n46. So this situation changes the shape of Ba having two squares over and under the six horizontal planes.( See my STRUCTURE OF Ba-132..Ba-138 ). In the following diagram of La the additional p57n57 fills the blank positions with S =+1 near the p30n30 . However for symmetrical arrangements the p37n37 of the square of Ba is moved to fill the blank positions with S = +1 near n29p29. Note that the p37 and p57 give at the six plane 2n of negative spins. Under this condition the number N of blank positions is given by The p39n39 gives 3n of negative spins The square of p38n38 and p40n40 gives 4n of positive spins The first horizontal plane gives 2(n) of positive spins The sixth horizontal plane gives 2(n) + 2n of negative spins The second and fifth plane give 4{n} + 8n of opposite spins The third plane gives 3(n) of positive spins The fourth plane gives 2(n) of negative spins That is N = 3n + 4n + 2(n) + 2(n) + 2n +4{n} + 8n + 3(n) +2(n) = 30 blank positions Since the La-139 of 25 extra neutrons has S = +7/2 we conclude that it is due to the S =+2 of the p37n37 and p57n57 and the S= +3/2 of the extra neutons by addiing 14 extra neutrons of positive spins and 11 extra neutrons of negative spins. That is S = +2 + 2{+1/2) + 8[ +1/2] + 4(+1/2) + 2{-1/2} + 9-1/2 ' ' DIAGRAM OF La-139 FORMING 30 BLANK POSITIONS Here the np7n57 is shown near the n30 p30 Also the p37n37 is shown at the symmetrical place near the n29p29 . But you cannot see the additional p49n49 the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' n40.......p40.......n ' ' n........p38..........n38 H. Square with n' ' n31………p12.........n12.......p32........n' ' n........p31........n11.........p11…… n32 Sixth H. plane' ' n37........ p29.........n10.........p10…… n30..........p57' ' p37......n29………..p9..........n9 …….p30..........n57 Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24...........n39' ' n........p23……..n3………p3………..n24............p39 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22.............n First H. plane' ' n39........p39.......n n39p39 with n' ' ' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' ' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE Here the n near the p14 fills the blank position formed by p51 and p14. While the {n} near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4.........n4.........p24......n39' n.......p23........n3........p3.........n24......p39 ' {n}...... p13......n13 n ' ' n ' ' ' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' '''Here the p53, np4, n55, p56 and n57 fill the blank positions of (n) and (p). So you can see the existing blank positions for receiving 2(n).and 2(p) ' (n)........p56....... n50.......p51......(n) ' ' p53........n42........p16......n16......p44.........n54''' ' n47........p25........n6........p6........n26.........p48' ' p45........n25........p5........n5........p26........ n46' ' n55.........p41.......n15.......p15.......n43' ' (n)........p49.......n52' Category:Fundamental physics concepts